Problem: $f(v, w) = (vw^2, \sin(v^2), v)$ What is $\dfrac{\partial f}{\partial w}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(w^2, 2v\cos(v^2), 1)$ (Choice B) B $(2w, 0, 0)$ (Choice C) C $(w^2 + 2vw, 2v\cos(v^2), 1)$ (Choice D) D $(2vw, 0, 0)$
Solution: The partial derivative of a vector valued function is component-wise partial differentiation. $\begin{aligned} &f(v, w) = (f_0(v, w), f_1(v, w), f_2(v, w)) \\ \\ &f_v = \left( \dfrac{\partial f_0}{\partial v}, \dfrac{\partial f_1}{\partial v}, \dfrac{\partial f_2}{\partial v} \right) \\ \\ &f_w = \left( \dfrac{\partial f_0}{\partial w}, \dfrac{\partial f_1}{\partial w}, \dfrac{\partial f_2}{\partial w} \right) \end{aligned}$ Because we're taking a partial derivative with respect to $w$, we'll treat $v$ as if it were a constant. Therefore, $f_w = (2vw, 0, 0)$.